UVA10882 Koerner's Pub【数学】

“Homer no function beer well without.”
Homer Simpson
Every Friday night, Alice, Clara and Mary go to Koerner’s pub to relax after a long week. At the pub, lots of guys ask them for their phone numbers. In fact, the three ladies are so popular that they have started counting the number of times each one is asked for her phone number during one evening. On day i, Alice, Clara and Mary were asked Ai , Ci and Mi times, respectively. 100 Fridays have passed, and the records were lost, but there are 3 things Alice still remembers.

X of the 100 days, Ai was equal to Ci.
Y of the 100 days, Ci was equal to Mi.
Z of the 100 days, Ai was equal to Mi.

How many time were Ai , Ci and Mi all equal?
Input
The first line of input gives the number of cases, N. N test cases follow. Each one contains the integers X, Y and Z on a line.
Output
For each test case, output one line containing ‘Case #x:’ followed by ‘Between A and B times.’, where A and B are the lowest and highest number of times that Ai , Ci and Mi could have been equal.
If the situation is impossible, print ‘The records are faulty.’ instead.
Sample Input
5
50 50 50
100 100 99
100 50 50
101 50 50
100 100 100
Sample Output
Case #1: Between 25 and 50 times.
Case #2: The records are faulty.
Case #3: Between 50 and 50 times.
Case #4: The records are faulty.
Case #5: Between 100 and 100 times.

问题链接：UVA10882 Koerner’s Pub
问题简述：（略）
问题分析：
给程序代码，暂时不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10882 Koerner's Pub */#include <bits/stdc++.h>using namespace std;int main()
{int t;scanf("%d", &t);for(int k = 1; k <= t; k++) {int a[3];scanf("%d%d%d", &a[0], &a[1], &a[2]);sort(a, a + 3);int ol = a[1] + a[2] - 100;ol = ol < 0 ? 0 : ol;int ra = 100 - (a[1] + a[2]);ra = ra < 0 ? 0 : ra;int ne = a[0] - ol - ra;ne = ne < 0 ? 0 : ne;if(ol > a[0])printf("Case #%d: The records are faulty.\n", k);elseprintf("Case #%d: Between %d and %d times.\n", k, ol + (ne + 1) / 2, a[0]);}return 0;
}

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